The following system of linear congruences in its given form cannot be solved
using the Chinese Remainder Theorem. Can you help me transform the system sufficiently
such that the Chinese Remainder Theorem can be applied, without actually solving the system?

I think it is difficult not to see a solution in the effort to transform the system! But there may be more than one solution.
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PADDec 27 '12 at 22:39

I know that x=16 works, but I am trying to reduce the system to modulo 3,5 and 7 rather than 15,21 and 35 because gcd(15,21,35) is not equal to 1. Is this a correct approach?
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flamingohatsDec 27 '12 at 22:46

Yes. That is correct. It works because in both case $x\equiv 1 \ \ (3)$. So, you have consistency. If the same happens on the third equation then you are o.k. I think it works there too because the answer is aganin $16$.
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PADDec 27 '12 at 22:52